| Date | May 2016 | Marks available | 6 | Reference code | 16M.2.hl.TZ0.4 |
| Level | HL | Paper | 2 | Time zone | TZ0 |
| Command term | Explain, Outline, and State | Question number | 4 | Adapted from | N/A |
Question
Tin(II) chloride is a white solid that is commonly used as a reducing agent.
(i) State why you would expect tin(II) chloride to have a similar lattice enthalpy to strontium chloride, using section 9 of the data booklet.
(ii) Calculate the molar enthalpy change when strontium chloride is dissolved in water, using sections 18 and 20 of the data booklet.
(iii) Tin(II) chloride reacts with water to precipitate the insoluble basic chloride, Sn(OH)Cl.
Suggest why tin(II) chloride is usually dissolved in dilute hydrochloric acid.
Tin can also exist in the +4 oxidation state.
Vanadium can be reduced from an oxidation state of +4 to +3 according to the equation:
(i) Calculate the cell potential, EΘ, and the standard free energy, ΔGΘ, change for the reaction between the VO2+ and Sn2+ ions, using sections 1 and 2 of the data booklet.
EΘ:
ΔGΘ:
(ii) Deduce, giving your reason, whether a reaction between Sn2+(aq) and VO2+(aq) would be spontaneous.
Outline, giving the full electron configuration of the vanadium atom, what is meant by the term transition metal.
In an aqueous solution of vanadium(III) chloride, the vanadium exists as [V (H2O)6]3+, [VCl (H2O)5]2+ or [VCl2(H2O)4]+ depending on the concentration of chloride ions in the solution.
(i) Describe how Cl− and H2O bond to the vanadium ion.
(ii) Outline what would happen to the wavelength at which the vanadium complex ions would absorb light as the water molecules are gradually replaced by chloride ions, using section 15 of the data booklet.
Eight successive ionisation energies of vanadium are shown in the graph below:
(i) State the sub-levels from which each of the first four electrons are lost.
First: Second: Third: Fourth:
(ii) Outline why there is an increase in ionization energy from electron 3 to electron 5.
(iii) Explain why there is a large increase in the ionization energy between electrons 5 and 6.
(iv) Vanadium is comprised almost entirely of 51V. State the number of neutrons an atom of 51V has in its nucleus.
Markscheme
(i)
same charge AND same/similar ionic radius
(ii)
enthalpy of hydration «= −1483 + 2 (−359)» = −2201 «kJmol−1»
enthalpy of solution «= 2170 − 2201» = −31 «kJmol−1»
Award [2] for correct final answer.
Award [1 max] for +31 «kJmol−1».
Award [1 max] for ±4371.
(iii)
hydrochloric acid shifts equilibrium to left
OR
hydrochloric acid prevents the basic chloride forming/precipitating
Accept “hydrochloric acid reacts with «basic» chloride” OR “hydrochloric acid suppresses salt hydrolysis”.
(i)
EΘ «= 0.34 − 0.15» = 0.19«V»
∆GOΘ«= − nFEΘ = −2 × 96500 × 0.19» = −36670 / −37000«J» / − 37«kJ»
Accept −18335 «J» or −18 «kJ» as equation not specified.
(ii)
yes AND ∆GΘ is negative
OR
yes AND EΘ for the cell is positive
OR
yes AND Sn2+ (aq) is a stronger reducing agent than V3+(aq)
OR
yes AND EΘ SN4+ (aq) is more negative that EΘ or VO2+ (aq)
OR
yes AND VO2+ (aq) is a stronger oxidizing agent than Sn4+ (aq)
OR
yes AND EΘ for VO2+ (aq) is more positive than EΘ for SN4+ (aq)
Do not accept reference to anti-clockwise rule.
1s22s22p63s23p63d34s2
OR
1s22s22p63s23p64s23d3
incomplete d «sub-» level/orbital/shell «in its compounds/ions»
(i)
give/donate a lone/non-bonding electron pair
Accept “through the formation of a dative/ coordinate bond”.
Accept “by acting as Lewis bases”.
Do not accept “act as ligands”.
(ii)
«more chlorido ligands» smaller energy gap between split d-orbitals
OR
Cl− is lower than H2O in spectrochemical series
OR
Cl− is a weaker ligand/has lower charge density
the absorption will move to longer wavelengths
OR
the absorption wavelength will increase
Do not accept answers in terms of change of frequency.
(i)
First: 4s AND Second: 4s AND Third: 3d AND Fourth: 3d
Do not apply ECF from (c).
(ii)
«in the same sub-shell and a» decrease in electron-electron repulsion
OR
«in the same sub-shell and» as more electrons removed, the pull of of the nucleus/positive ions holds the remaining electrons more tightly
Do not accept “greater nuclear charge/ effective nuclear charge”.
(iii)
electron 5 is lost from the 3d orbital
OR
electron 5 is lost from the valence shell
electron 6 is lost from a 3p orbital
OR
electron 6 is lost from a «complete» inner shell
3p orbital/complete inner shell experiences a much larger effective nuclear charge
OR
3p orbital/complete inner shell is less well shielded
OR
3p orbital/complete inner shell is nearer the nucleus
Award [1 max] (for M1/M2) (ECF) if candidate recognises electrons 5 and 6 are from different levels.
(iv)
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